Answer
$\dfrac{2}{k}$
Work Step by Step
We know that $\sin x$ is a periodic function with period $\dfrac{2\pi}{k}$.
Now, $A=\int_0^{\pi/k} \sin kx dx$
This implies that
$[(\dfrac{-1}{k}) \cos kx]_0^{\pi/k}=\dfrac{-1}{k}\cos k(1/k)-(\dfrac{-1}{k}\cos 0)$
Thus, $A=\dfrac{2}{k}$